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部分分数分解計算機
部分分数分解を段階的に求める
このオンライン計算機は、有理関数の部分分数分解を、手順を示しながら求めます。
Solution
Your input: perform the partial fraction decomposition of $$$\frac{1}{\left(\left(2 n + 1\right)^{2} - 4\right)^{2}}$$$
Factor the denominator: $$$\frac{1}{\left(\left(2 n + 1\right)^{2} - 4\right)^{2}}=\frac{1}{\left(2 n - 1\right)^{2} \left(2 n + 3\right)^{2}}$$$
The form of the partial fraction decomposition is
$$\frac{1}{\left(2 n - 1\right)^{2} \left(2 n + 3\right)^{2}}=\frac{A}{2 n - 1}+\frac{B}{\left(2 n - 1\right)^{2}}+\frac{C}{2 n + 3}+\frac{D}{\left(2 n + 3\right)^{2}}$$
Write the right-hand side as a single fraction:
$$\frac{1}{\left(2 n - 1\right)^{2} \left(2 n + 3\right)^{2}}=\frac{\left(2 n - 1\right)^{2} \left(2 n + 3\right) C + \left(2 n - 1\right)^{2} D + \left(2 n - 1\right) \left(2 n + 3\right)^{2} A + \left(2 n + 3\right)^{2} B}{\left(2 n - 1\right)^{2} \left(2 n + 3\right)^{2}}$$
The denominators are equal, so we require the equality of the numerators:
$$1=\left(2 n - 1\right)^{2} \left(2 n + 3\right) C + \left(2 n - 1\right)^{2} D + \left(2 n - 1\right) \left(2 n + 3\right)^{2} A + \left(2 n + 3\right)^{2} B$$
Expand the right-hand side:
$$1=8 n^{3} A + 8 n^{3} C + 20 n^{2} A + 4 n^{2} B + 4 n^{2} C + 4 n^{2} D + 6 n A + 12 n B - 10 n C - 4 n D - 9 A + 9 B + 3 C + D$$
Collect up the like terms:
$$1=n^{3} \left(8 A + 8 C\right) + n^{2} \left(20 A + 4 B + 4 C + 4 D\right) + n \left(6 A + 12 B - 10 C - 4 D\right) - 9 A + 9 B + 3 C + D$$
The coefficients near the like terms should be equal, so the following system is obtained:
$$\begin{cases} 8 A + 8 C = 0\\20 A + 4 B + 4 C + 4 D = 0\\6 A + 12 B - 10 C - 4 D = 0\\- 9 A + 9 B + 3 C + D = 1 \end{cases}$$
Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{1}{32}$$$, $$$B=\frac{1}{16}$$$, $$$C=\frac{1}{32}$$$, $$$D=\frac{1}{16}$$$
Therefore,
$$\frac{1}{\left(2 n - 1\right)^{2} \left(2 n + 3\right)^{2}}=\frac{- \frac{1}{32}}{2 n - 1}+\frac{\frac{1}{16}}{\left(2 n - 1\right)^{2}}+\frac{\frac{1}{32}}{2 n + 3}+\frac{\frac{1}{16}}{\left(2 n + 3\right)^{2}}$$
Answer: $$$\frac{1}{\left(\left(2 n + 1\right)^{2} - 4\right)^{2}}=\frac{- \frac{1}{32}}{2 n - 1}+\frac{\frac{1}{16}}{\left(2 n - 1\right)^{2}}+\frac{\frac{1}{32}}{2 n + 3}+\frac{\frac{1}{16}}{\left(2 n + 3\right)^{2}}$$$